At the extreme values of a function f(x), the first derivative f'(x) = 0

f(x) = 3x^2 - 4x + 5

f'(x) = 6x - 4

6x - 4 = 0

=> x = 4/6

=> x = 2/3

The extreme value of the function is f(2/3) = 3*4/9 - 4*2/3 + 5

=> 4/3 - 8/3 + 5

=> 11/3

**The extreme value of the function is 11/3**

Posted on

f(x)= 3x^2 - 4x + 5

we need to find the extreme value of f(x).

First we will find the critical points.

==. f'(x) = 6x -4 = 0

==> x = 4/6 = 2/3

Then the function f(x) has an extreme value at f(2/3)

==> f(2/3) = 3*(2/3)^2 - 4*2/3 + 5

==> f(2/3) = 12/9 - 8/3 + 5 = ( 12 - 24 + 45) / 9 = 33/9 = 11/3

**Then the function has extreme value at the point f(2/3) = 11/3 OR the point ( 2/3, 11/3) **

Posted on

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